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Thread: convergent sequences

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    Last edited by britnumber; Oct 9th 2008 at 08:34 PM.
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    Quote Originally Posted by britnumber View Post
    Let X=(x_n) be a sequence of positive real numbers such that = L >1. Show that X is not a bounded sequence and hence is not convergent
    Let $\displaystyle \epsilon > 0$ so that $\displaystyle L - \epsilon > 1$.

    And therefore there is $\displaystyle N$ so that if $\displaystyle n\geq N$ then $\displaystyle \left| \frac{x_{n+1}}{x_n} - L \right| < \epsilon$.

    Therefore, $\displaystyle L - \epsilon< \frac{x_{n+1}}{x_n} \implies (L - \epsilon)x_n < x_{n+1}$.

    Therefore,
    $\displaystyle (L - \epsilon)x_N < x_{N+1}$
    $\displaystyle (L - \epsilon)x_{N+1} < x_{N+2}$
    $\displaystyle (L - \epsilon)x_{N+2} < x_{N+3}$
    ...
    Thus,
    $\displaystyle (L - \epsilon)^2x_N < x_{N+2}$
    $\displaystyle (L - \epsilon)^3x_N < x_{N+3}$
    ...
    In general,
    $\displaystyle (L - \epsilon)^kx_N < x_{N+k}$

    But $\displaystyle (L - \epsilon)^k$ can be made arbitary large since $\displaystyle L - \epsilon > 1$.
    Therefore, $\displaystyle \{ x_n\}$ is unbounded.
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